Class 12: Areas of Bounded Regions – Exercise 21.1 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

$\displaystyle \textbf{Using integration, find the area of the following regions:}$

$\displaystyle \textbf{Question 1: }\text{Using integration, find the area of the region bounded between the line } \\ x=2\text{ and the parabola }y^{2}=8x.$
$\displaystyle \text{Answer:}$

$\displaystyle y^2=8x \text{ represents a parabola with vertex at the origin and axis of symmetry along} \\ \text{the positive direction of the } x\text{-axis}$
$\displaystyle x=2 \text{ is a line parallel to the } y\text{-axis}$
$\displaystyle \text{Let } (x,y) \text{ be a point on the parabola } y^2=8x$
$\displaystyle \text{Since the parabola } y^2=8x \text{ is symmetric about the } x\text{-axis}$
$\displaystyle \therefore \text{Required area } = 2(\text{area above the } x\text{-axis})$
$\displaystyle \text{On slicing the area above the } x\text{-axis into vertical strips of length } |y| \text{ and width } dx$
$\displaystyle \Rightarrow \text{Area of rectangular strip } = |y|\,dx$
$\displaystyle \text{The approximating rectangle moves between } x=0 \text{ and } x=2$
$\displaystyle \Rightarrow A=2\int_0^2 |y|\,dx=2\int_0^2 y\,dx \text{ since } y>0$
$\displaystyle =2\int_0^2 \sqrt{8x}\,dx$
$\displaystyle =2\int_0^2 2\sqrt{2x}\,dx$
$\displaystyle =4\sqrt{2}\int_0^2 \sqrt{x}\,dx$
$\displaystyle =4\sqrt{2}\left[\frac{2}{3}x^{3/2}\right]_0^2$
$\displaystyle =\frac{8}{3}\sqrt{2}\left[2^{3/2}-0\right]$
$\displaystyle =\frac{8}{3}\cdot2^2=\frac{32}{3}$
$\displaystyle \therefore \text{Area } A=\frac{32}{3}\text{ sq. units}$

$\displaystyle \textbf{Question 2: }\text{Using integration, find the area of the region bounded by the line } \\ y-1=x,\ \text{the }x\text{-axis and the ordinates }x=-2\text{ and }x=3.$
$\displaystyle \text{Answer:}$

$\displaystyle \text{The line } y-1=x \text{ cuts the } x\text{-axis at } D(-1,0) \text{ and the } y\text{-axis at } A(0,1)$
$\displaystyle \text{The lines } x=-2 \text{ and } x=3 \text{ are straight lines parallel to the } y\text{-axis}$
$\displaystyle \text{Since } |y|=-(1+x) \text{ for } x\le -1$
$\displaystyle \text{and } |y|=(1+x) \text{ for } x>-1$
$\displaystyle \text{Area of the region bounded by the line } y-1=x,\ x\text{-axis and the ordinates } x=-2 \text{ and } x=3$
$\displaystyle \Rightarrow A=\int_{-2}^{3}|y|\,dx$
$\displaystyle =\int_{-2}^{-1}|y|\,dx+\int_{-1}^{0}|y|\,dx+\int_{0}^{3}|y|\,dx$
$\displaystyle =\int_{-2}^{-1}-(1+x)\,dx+\int_{-1}^{0}(1+x)\,dx+\int_{0}^{3}(1+x)\,dx$
$\displaystyle =\left[-x-\frac{x^2}{2}\right]_{-2}^{-1}+\left[x+\frac{x^2}{2}\right]_{-1}^{0}+\left[x+\frac{x^2}{2}\right]_{0}^{3}$
$\displaystyle =\left[1-\frac{1}{2}-2+\frac{4}{2}\right]+\left[0+0+1-\frac{1}{2}\right]+\left[3+\frac{9}{2}\right]$
$\displaystyle =\left(-1+\frac{3}{2}\right)+\left(1-\frac{1}{2}\right)+\left(3+\frac{9}{2}\right)$
$\displaystyle =3+\frac{3-1+9}{2}=3+\frac{11}{2}=\frac{17}{2}$
$\displaystyle \therefore \text{Area of the bounded region }=\frac{17}{2}\text{ sq. units}$

$\displaystyle \textbf{Question 3: }\text{Find the area of the region bounded by the parabola } y^{2}=4ax \\ \text{ and the line }x=a.$
$\displaystyle \text{Answer:}$

$\displaystyle \text{The equation } y^2=4ax \text{ represents a parabola with vertex at } (0,0) \text{ and } \\ \text{symmetric about the positive direction of the } x\text{-axis}$
$\displaystyle x=a \text{ is a line parallel to the } y\text{-axis and cuts the } x\text{-axis at } (a,0)$
$\displaystyle \text{Vertical strips of length } |y| \text{ and width } dx \text{ are taken in the quadrant OLSO}$
$\displaystyle \text{Area of approximating rectangle } = |y|\,dx$
$\displaystyle \text{Since the rectangle moves between } x=0 \text{ and } x=a$
$\displaystyle \text{and the parabola is symmetric about the } x\text{-axis}$
$\displaystyle \text{Required shaded area OLSO } = A = 2\times\text{Area OLMO}$
$\displaystyle A=2\int_0^a |y|\,dx=2\int_0^a y\,dx \text{ since } y>0 \Rightarrow |y|=y$
$\displaystyle =2\int_0^a \sqrt{4ax}\,dx$
$\displaystyle =2\int_0^a 2\sqrt{ax}\,dx$
$\displaystyle =4\int_0^a \sqrt{ax}\,dx$
$\displaystyle =4\sqrt{a}\int_0^a \sqrt{x}\,dx$
$\displaystyle =4\sqrt{a}\left[\frac{2}{3}x^{3/2}\right]_0^a$
$\displaystyle =\frac{8}{3}\sqrt{a}\,[a^{3/2}-0]$
$\displaystyle =\frac{8}{3}a^2$
$\displaystyle \therefore A=\frac{8}{3}a^2\text{ sq. units}$

$\displaystyle \textbf{Question 4: }\text{Find the area lying above the }x\text{-axis and under the parabola } \\ y=4x-x^{2}.$
$\displaystyle \text{Answer:}$

$\displaystyle \text{The equation } y=4x-x^2 \text{ represents a parabola opening downwards and cutting } \\ \text{the } x\text{-axis at } O(0,0) \text{ and } B(4,0)$
$\displaystyle \text{On slicing the region above the } x\text{-axis into vertical strips of length } |y| \text{ and width } dx$
$\displaystyle \text{Area of the corresponding rectangular strip } = |y|\,dx$
$\displaystyle \text{Since the rectangle moves from } x=0 \text{ to } x=4$
$\displaystyle \text{Required area OABO is }$
$\displaystyle A=\int_0^4 |y|\,dx=\int_0^4 y\,dx \text{ since } y>0 \text{ for } 0\le x\le4 \Rightarrow |y|=y$
$\displaystyle =\int_0^4 (4x-x^2)\,dx$
$\displaystyle =\left[\frac{4x^2}{2}-\frac{x^3}{3}\right]_0^4$
$\displaystyle =\left[32-\frac{64}{3}\right]$
$\displaystyle =\frac{32}{3}$
$\displaystyle \therefore \text{Area }=\frac{32}{3}\text{ sq. units}$

$\displaystyle \textbf{Question 5: }\text{Draw a rough sketch to indicate the region bounded between the curve } \\ y^{2}=4x\text{ and the line }x=3.\ \text{Also, find the area of this region.}$
$\displaystyle \text{Answer:}$

$\displaystyle y^2=4x \text{ represents a parabola with vertex at } (0,0) \text{ and axis of symmetry along the } \\ \text{positive direction of the } x\text{-axis}$
$\displaystyle x=3 \text{ is a line parallel to the } y\text{-axis and cuts the } x\text{-axis at } (3,0)$
$\displaystyle \text{Since the parabola } y^2=4x \text{ is symmetrical about the } x\text{-axis}$
$\displaystyle \therefore \text{Required area } A = 2\times(\text{area above the } x\text{-axis})$
$\displaystyle \text{On slicing the area above the } x\text{-axis into vertical strips of length } |y| \text{ and width } dx$
$\displaystyle \text{Area of the corresponding rectangular strip } = |y|\,dx$
$\displaystyle \text{The corresponding rectangle moves from } x=0 \text{ to } x=3$
$\displaystyle A=2\int_0^3 |y|\,dx=2\int_0^3 y\,dx \text{ since } y>0 \Rightarrow |y|=y$
$\displaystyle =2\int_0^3 \sqrt{4x}\,dx$
$\displaystyle =2\int_0^3 2\sqrt{x}\,dx$
$\displaystyle =4\int_0^3 \sqrt{x}\,dx$
$\displaystyle =4\left[\frac{2}{3}x^{3/2}\right]_0^3$
$\displaystyle =\frac{8}{3}\left[3^{3/2}-0\right]$
$\displaystyle =\frac{8}{3}\cdot3\sqrt{3}=8\sqrt{3}$
$\displaystyle \therefore \text{Area of the region bounded by } y^2=4x \text{ and } x=3 \text{ is } 8\sqrt{3}\text{ sq. units}$

$\displaystyle \textbf{Question 6: }\text{Make a rough sketch of the graph of the function } y=4-x^{2},\ \\ 0\le x\le 2 \text{ and determine the area enclosed by the curve, the }x\text{-axis and the lines } \\ x=0\text{ and }x=2.$
$\displaystyle \text{Answer:}$

$\displaystyle y=4-x^2,\ 0\le x\le2 \text{ represents a half parabola with vertex at } (4,0)$
$\displaystyle x=2 \text{ represents a line parallel to the } y\text{-axis and cuts the } x\text{-axis at } (2,0)$
$\displaystyle \text{In quadrant OABO, consider a vertical strip of length } |y| \text{ and width } dx$
$\displaystyle \text{Area of the approximating rectangle } = |y|\,dx$
$\displaystyle \text{The approximating rectangle moves from } x=0 \text{ to } x=2$
$\displaystyle A=\text{Area OABO}=\int_0^2 |y|\,dx$
$\displaystyle =\int_0^2 y\,dx \text{ since } y>0 \Rightarrow |y|=y$
$\displaystyle =\int_0^2 (4-x^2)\,dx$
$\displaystyle =\left[4x-\frac{x^3}{3}\right]_0^2$
$\displaystyle =8-\frac{8}{3}$
$\displaystyle =\frac{16}{3}$
$\displaystyle \therefore \text{The area enclosed by the curve, the } x\text{-axis and the given lines }=\frac{16}{3}\text{ sq. units}$

$\displaystyle \textbf{Question 7: }\text{Sketch the graph of }y=\sqrt{x+1}\text{ and determine the area of the region} \\ \text{enclosed by the curve, the } x\text{-axis and the lines }x=0,\ x=4.$
$\displaystyle \text{Answer:}$

$\displaystyle y=\sqrt{x+1}\text{ in }[0,4]\text{ represents a curve which is part of a parabola}$
$\displaystyle x=4 \text{ represents a line parallel to the } y\text{-axis and cuts the } x\text{-axis at }(4,0)$
$\displaystyle \text{The enclosed area bounded by the curve and the lines } x=0 \text{ and } x=4 \text{ is OABCO}$
$\displaystyle \text{Consider a vertical strip of length } |y| \text{ and width } dx$
$\displaystyle \text{Area of the approximating rectangle }=|y|\,dx$
$\displaystyle \text{The approximating rectangle moves from } x=0 \text{ to } x=4$
$\displaystyle A=\text{Area OABCO}=\int_0^4 |y|\,dx$
$\displaystyle =\int_0^4 y\,dx \text{ since } y>0 \Rightarrow |y|=y$
$\displaystyle =\int_0^4 \sqrt{x+1}\,dx$
$\displaystyle =\int_0^4 (x+1)^{1/2}\,dx$
$\displaystyle =\left[\frac{(x+1)^{3/2}}{3/2}\right]_0^4$
$\displaystyle =\frac{2}{3}\left[(5)^{3/2}-1\right]$
$\displaystyle \therefore \text{Enclosed area between the curve and the given lines }=\frac{2}{3}(5^{3/2}-1)\text{ sq. units}$

$\displaystyle \textbf{Question 8: }\text{Find the area under the curve }y=\sqrt{6x+4}\text{ above }x\text{-axis from } \\ x=0 \text{ to }x=2.\ \text{Draw a sketch of curve also.}$
$\displaystyle \text{Answer:}$

$\displaystyle y=\sqrt{6x+4} \text{ represents a parabola with vertex at } \left(-\frac{2}{3},0\right) \text{ and symmetrical about the } x\text{-axis}$
$\displaystyle x=0 \text{ is the } y\text{-axis. The curve cuts it at } A(0,2) \text{ and } A'(0,-2)$
$\displaystyle x=2 \text{ is a line parallel to the } y\text{-axis and cuts the } x\text{-axis at } C(2,0)$
$\displaystyle \text{The enclosed area of the curve between } x=0 \text{ and } x=2 \text{ and above the } x\text{-axis is area OABC}$
$\displaystyle \text{Consider a vertical strip of length } |y| \text{ and width } dx$
$\displaystyle \text{Area of the approximating rectangle }=|y|\,dx$
$\displaystyle \text{The approximating rectangle moves from } x=0 \text{ to } x=2$
$\displaystyle A=\text{area OABC}=\int_0^2 |y|\,dx$
$\displaystyle =\int_0^2 y\,dx \text{ since } y>0 \Rightarrow |y|=y$
$\displaystyle =\int_0^2 \sqrt{6x+4}\,dx$
$\displaystyle =\int_0^2 (6x+4)^{1/2}\,dx$
$\displaystyle =\left[\frac{(6x+4)^{3/2}}{6\cdot\frac{3}{2}}\right]_0^2$
$\displaystyle =\frac{2}{18}\left[(16)^{3/2}-(4)^{3/2}\right]$
$\displaystyle =\frac{2}{18}\left[64-8\right]$
$\displaystyle =\frac{2}{18}\times56=\frac{56}{9}$
$\displaystyle \therefore \text{Enclosed area }=\frac{56}{9}\text{ sq. units}$

$\displaystyle \textbf{Question 9: }\text{Draw the rough sketch of }y^{2}+1=x,\ x\le 2.\ \text{Find the area enclosed } \\ \text{by the curve and the line }x=2.$
$\displaystyle \text{Answer:}$

$\displaystyle y^2+1=x,\ x\le2 \text{ represents a parabola with vertex at } (1,0) \text{ and symmetrical about the } \\ \text{positive direction of the } x\text{-axis}$
$\displaystyle x=2 \text{ is a line parallel to the } y\text{-axis and cuts the } x\text{-axis at } (2,0)$
$\displaystyle \text{Consider a vertical section of height } |y| \text{ and width } dx \text{ in the first quadrant}$
$\displaystyle \text{Area of the corresponding rectangle } = |y|\,dx$
$\displaystyle \text{Since the rectangle moves from } x=1 \text{ to } x=2 \text{ and the curve is symmetrical}$
$\displaystyle \therefore A=\text{area ABCA}=2\times\text{area ABDA}$
$\displaystyle A=2\int_1^2 |y|\,dx$
$\displaystyle =2\int_1^2 y\,dx \text{ since } y>0 \Rightarrow |y|=y$
$\displaystyle =2\int_1^2 \sqrt{x-1}\,dx \text{ since } y^2+1=x \Rightarrow y=\sqrt{x-1}$
$\displaystyle =2\left[\frac{(x-1)^{3/2}}{3/2}\right]_1^2$
$\displaystyle =\frac{4}{3}\left[1^{3/2}-0\right]$
$\displaystyle =\frac{4}{3}$
$\displaystyle \therefore \text{Enclosed area }=\frac{4}{3}\text{ sq. units}$

$\displaystyle \textbf{Question 10: }\text{Draw a rough sketch of the graph of the curve }\frac{x^{2}}{4}+\frac{y^{2}}{9}=1 \\ \text{ and evaluate the area of the region under the curve and above }x\text{-axis.}$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Since in the given equation } \frac{x^2}{4}+\frac{y^2}{9}=1,\ \text{all the powers of } \\ x \text{ and } y \text{ are even, the curve is symmetrical about both the axes}$
$\displaystyle \therefore \text{Area enclosed by the curve and above the } x\text{-axis }=A'BA=2\times(\text{area } \\ \text{enclosed by the ellipse and the } x\text{-axis in the first quadrant})$
$\displaystyle (2,0),(-2,0)\text{ are the points of intersection of the curve and the } x\text{-axis}$
$\displaystyle (0,3),(0,-3)\text{ are the points of intersection of the curve and the } y\text{-axis}$
$\displaystyle \text{Slice the area in the first quadrant into vertical strips of height } |y| \text{ and width } dx$
$\displaystyle \text{Area of the approximating rectangle }=|y|\,dx$
$\displaystyle \text{The approximating rectangle moves between } x=0 \text{ and } x=2$
$\displaystyle A=\text{Area enclosed above the } x\text{-axis}=2\int_0^2 |y|\,dx$
$\displaystyle =2\int_0^2 y\,dx \text{ since } y>0 \Rightarrow |y|=y$
$\displaystyle =2\int_0^2 \frac{3}{2}\sqrt{4-x^2}\,dx$
$\displaystyle =3\int_0^2 \sqrt{4-x^2}\,dx$
$\displaystyle =3\left[\frac{1}{2}x\sqrt{4-x^2}+\frac{1}{2}4\sin^{-1}\!\left(\frac{x}{2}\right)\right]_0^2$
$\displaystyle =3\left[0+\frac{1}{2}\times4\times\frac{\pi}{2}\right]$
$\displaystyle =3\pi$
$\displaystyle \therefore \text{Area of the enclosed region above the } x\text{-axis }=3\pi\text{ sq. units}$

$\displaystyle \textbf{Question 11: }\text{Sketch the region }\{(x,y):9x^{2}+4y^{2}=36\}\text{ and find the area of the} \\ \text{region enclosed by it, using integration.}$
$\displaystyle \text{Answer:}$

$\displaystyle \text{We have,}$
$\displaystyle 9x^2+4y^2=36\ \text{......(1)}$
$\displaystyle \Rightarrow 4y^2=36-9x^2$
$\displaystyle \Rightarrow y^2=\frac{9}{4}(4-x^2)$
$\displaystyle \Rightarrow y=\frac{3}{2}\sqrt{4-x^2}\ \text{......(2)}$
$\displaystyle \text{From (1), we get } \frac{x^2}{4}+\frac{y^2}{9}=1$
$\displaystyle \text{Since in the given equation } \frac{x^2}{4}+\frac{y^2}{9}=1,\text{ all the powers of both }x\text{ and }y \\ \text{ are even, the curve is symmetrical about both the axes.}$
$\displaystyle \therefore \text{Area enclosed by the curve }=4\times\text{area enclosed by ellipse and }x\text{-axis in first quadrant.}$
$\displaystyle (2,0),(-2,0)\text{ are the points of intersection of curve and }x\text{-axis.}$
$\displaystyle (0,3),(0,-3)\text{ are the points of intersection of curve and }y\text{-axis.}$
$\displaystyle \text{Slicing the area in the first quadrant into vertical strips of height }|y|\text{ and width }dx.$
$\displaystyle \therefore \text{Area of approximating rectangle }=|y|dx.$
$\displaystyle \text{Approximating rectangle can move between }x=0\text{ and }x=2.$
$\displaystyle A=\text{Area enclosed curve }=4\int_0^2|y|\,dx$
$\displaystyle \Rightarrow A=4\int_0^2\frac{3}{2}\sqrt{4-x^2}\,dx\ \text{(From (2))}$
$\displaystyle =6\int_0^2\sqrt{4-x^2}\,dx$
$\displaystyle =6\left[\frac{x}{2}\sqrt{4-x^2}+\frac{4}{2}\sin^{-1}\frac{x}{2}\right]_0^2$
$\displaystyle =6\left[0+\frac{1}{2}\times4\sin^{-1}1\right]$
$\displaystyle =6\left[\frac{1}{2}\times4\times\frac{\pi}{2}\right]$
$\displaystyle \Rightarrow A=6\pi\ \text{sq. units}$

$\displaystyle \textbf{Question 12: }\text{Draw a rough sketch of the graph of the function } y=2\sqrt{1-x^{2}}, \\ x\in[0,1] \text{ and evaluate the area enclosed between the curve and the }x\text{-axis.}$
$\displaystyle \text{Answer:}$

$\displaystyle \text{We have}$
$\displaystyle y=2\sqrt{1-x^2}$
$\displaystyle \Rightarrow \frac{y}{2}=\sqrt{1-x^2}$
$\displaystyle \Rightarrow \frac{y^2}{4}=1-x^2$
$\displaystyle \Rightarrow x^2+\frac{y^2}{4}=1$
$\displaystyle \Rightarrow \frac{x^2}{1}+\frac{y^2}{4}=1$
$\displaystyle \text{Since in the given equation } \frac{x^2}{1}+\frac{y^2}{4}=1,\text{ all powers of }x\text{ and }y \\ \text{ are even, the curve is symmetrical about both the axes}$
$\displaystyle \therefore \text{Required area }=\text{area enclosed by ellipse and }x\text{-axis in first quadrant}$
$\displaystyle (1,0),(-1,0)\text{ are the points of intersection of the curve and }x\text{-axis}$
$\displaystyle (0,2),(0,-2)\text{ are the points of intersection of the curve and }y\text{-axis}$
$\displaystyle \text{Slicing the area in the first quadrant into vertical strips of height }|y|\text{ and width }dx$
$\displaystyle \text{Area of approximating rectangle }=|y|\,dx$
$\displaystyle \text{Approximating rectangle moves between }x=0\text{ and }x=1$
$\displaystyle A=\text{Area of enclosed curve above }x\text{-axis}=\int_0^1 |y|\,dx$
$\displaystyle =\int_0^1 2\sqrt{1-x^2}\,dx$
$\displaystyle =2\int_0^1 \sqrt{1-x^2}\,dx$
$\displaystyle =2\left[\frac{x}{2}\sqrt{1-x^2}+\frac{1}{2}\sin^{-1}x\right]_0^1$
$\displaystyle =2\left[\frac{1}{2}\sin^{-1}1\right]$
$\displaystyle =2\left[\frac{1}{2}\left(\frac{\pi}{2}-0\right)\right]$
$\displaystyle =\frac{\pi}{2}$
$\displaystyle \therefore \text{Enclosed area }=\frac{\pi}{2}\text{ sq. units}$

$\displaystyle \textbf{Question 13: }\text{Determine the area under the curve }y=\sqrt{a^{2}-x^{2}} \text{ included between} \\ \text{the lines }x=0\text{ and }x=a.$
$\displaystyle \text{Answer:}$

$\displaystyle \text{We have}$
$\displaystyle y=\sqrt{a^2-x^2}$
$\displaystyle \Rightarrow y^2=a^2-x^2$
$\displaystyle \Rightarrow x^2+y^2=a^2$
$\displaystyle \text{Since in the given equation } x^2+y^2=a^2,\text{ all powers of }x\text{ and }y \\ \text{ are even, the curve is symmetrical about both the axes}$
$\displaystyle \therefore \text{Required area }=\text{area enclosed by the circle in the first quadrant}$
$\displaystyle (a,0),(-a,0)\text{ are the points of intersection of the curve and }x\text{-axis}$
$\displaystyle (0,a),(0,-a)\text{ are the points of intersection of the curve and }y\text{-axis}$
$\displaystyle \text{Slicing the area in the first quadrant into vertical strips of height }|y|\text{ and width }dx$
$\displaystyle \text{Area of approximating rectangle }=|y|\,dx$
$\displaystyle \text{Approximating rectangle moves between }x=0\text{ and }x=a$
$\displaystyle A=\text{Area of enclosed curve in the first quadrant}=\int_0^a |y|\,dx$
$\displaystyle =\int_0^a \sqrt{a^2-x^2}\,dx$
$\displaystyle =\left[\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\!\left(\frac{x}{a}\right)\right]_0^a$
$\displaystyle =\frac{a^2}{2}\sin^{-1}1$
$\displaystyle =\frac{a^2}{2}\cdot\frac{\pi}{2}$
$\displaystyle =\frac{a^2\pi}{4}$
$\displaystyle \therefore \text{Enclosed area }=\frac{a^2\pi}{4}\text{ sq. units}$

$\displaystyle \textbf{Question 14: }\text{Using integration, find the area of the region bounded by the line } \\ 2y=5x+7,\ x\text{-axis and the lines }x=2\text{ and }x=8.$
$\displaystyle \text{Answer:}$

$\displaystyle \text{We have}$
$\displaystyle \text{Straight line }2y=5x+7\text{ intersects the }x\text{-axis and }y\text{-axis at }(-1.4,0)\text{ and }(0,3.5)\text{ respectively}$
$\displaystyle x=2 \text{ and } x=8 \text{ are straight lines as shown in the figure}$
$\displaystyle \text{The shaded region is the required region whose area is to be found}$
$\displaystyle \text{When the shaded region is sliced into vertical strips, each strip has its lower end on the } \\ x\text{-axis and upper end on the line }2y=5x+7$
$\displaystyle \text{Hence, the approximating rectangle has length }y\text{ and width }dx\text{ and area }=y\,dx$
$\displaystyle \text{The approximating rectangle moves from }x=2\text{ to }x=8$
$\displaystyle \therefore A=\int_2^8 y\,dx$
$\displaystyle =\int_2^8 \frac{5x+7}{2}\,dx$
$\displaystyle =\frac{1}{2}\int_2^8 (5x+7)\,dx$
$\displaystyle =\frac{1}{2}\left[\frac{5}{2}x^2+7x\right]_2^8$
$\displaystyle =\frac{1}{2}\left[\frac{5}{2}(64-4)+7(8-2)\right]$
$\displaystyle =\frac{1}{2}\left[150+42\right]$
$\displaystyle =\frac{1}{2}\times192$
$\displaystyle =96$
$\displaystyle \therefore \text{Area of the shaded region }=96\text{ sq. units}$

$\displaystyle \textbf{Question 15: }\text{Using definite integrals, find the area of the circle } \\ x^{2}+y^{2}=a^{2}.$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Area of the circle }x^2+y^2=a^2\text{ is }4\text{ times the area enclosed between }x=0\text{ and }x=a\text{ in the first quadrant}$
$\displaystyle A=4\int_0^a |y|\,dx$
$\displaystyle =4\int_0^a \sqrt{a^2-x^2}\,dx$
$\displaystyle =4\left[\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\!\left(\frac{x}{a}\right)\right]_0^a$
$\displaystyle =4\left[0+\frac{a^2}{2}\sin^{-1}1\right]$
$\displaystyle =4\left[\frac{a^2}{2}\cdot\frac{\pi}{2}\right]$
$\displaystyle =a^2\pi$
$\displaystyle \therefore \text{Area of the circle }=a^2\pi\text{ sq. units}$

$\displaystyle \textbf{Question 16: }\text{Using integration, find the area of the region bounded by the } \\ \text{following curves, after making a rough sketch: }y=1+\lvert x+1\rvert,\ \\ x=-2,\ x=3,\ y=0.$
$\displaystyle \text{Answer:}$

$\displaystyle \text{We have}$
$\displaystyle y=1+|x+1|\text{ intersects }x=-2\text{ at }(-2,2)\text{ and }x=3\text{ at }(3,5)$
$\displaystyle y=0\text{ is the }x\text{-axis}$
$\displaystyle \text{The shaded region is the required region whose area is to be found}$
$\displaystyle y=1+|x+1|$
$\displaystyle =\begin{cases}1-(x+1),&x<-1\\[4pt]1+(x+1),&x\ge -1\end{cases}$
$\displaystyle =\begin{cases}-x,&x<-1\\[4pt]x+2,&x\ge -1\end{cases}$
$\displaystyle \text{Let the required area be }A.\ \text{Since limits on }x\text{ are given, we use vertical strips}$
$\displaystyle A=\int_{-2}^{3}|y|\,dx$
$\displaystyle =\int_{-2}^{-1}|y|\,dx+\int_{-1}^{3}|y|\,dx$
$\displaystyle =\int_{-2}^{-1}(-x)\,dx+\int_{-1}^{3}(x+2)\,dx$
$\displaystyle =\left[-\frac{x^{2}}{2}\right]_{-2}^{-1}+\left[\frac{x^{2}}{2}+2x\right]_{-1}^{3}$
$\displaystyle =\left[-\frac{1}{2}+\frac{4}{2}\right]+\left[\frac{9}{2}+6-\frac{1}{2}+2\right]$
$\displaystyle =\frac{3}{2}+\left[8+4\right]$
$\displaystyle =\frac{27}{2}$
$\displaystyle \therefore \text{Area of the shaded region }=\frac{27}{2}\text{ sq. units}$

$\displaystyle \textbf{Question 17: }\text{Sketch the graph }y=\lvert x-5\rvert.\ \text{Evaluate }\int_{0}^{1}\lvert x-5\rvert\,dx.\ \\ \text{What does this value of the integral represent on the graph.}$
$\displaystyle \text{Answer:}$

$\displaystyle \text{We have}$
$\displaystyle y=|x-5|\text{ intersects }x=0\text{ and }x=1\text{ at }(0,5)\text{ and }(1,4)\text{ respectively}$
$\displaystyle \text{Now}$
$\displaystyle y=|x-5|$
$\displaystyle =-(x-5)\text{ for all }x\in(0,1)$
$\displaystyle \text{Integration represents the area enclosed by the graph from }x=0\text{ to }x=1$
$\displaystyle A=\int_0^1 |y|\,dx$
$\displaystyle =\int_0^1 |x-5|\,dx$
$\displaystyle =\int_0^1 -(x-5)\,dx$
$\displaystyle =-\int_0^1 (x-5)\,dx$
$\displaystyle =-\left[\frac{x^2}{2}-5x\right]_0^1$
$\displaystyle =-\left[\left(\frac{1}{2}-5\right)-(0-0)\right]$
$\displaystyle =-\left(-\frac{9}{2}\right)$
$\displaystyle =\frac{9}{2}$
$\displaystyle \therefore \text{Area enclosed }=\frac{9}{2}\text{ sq. units}$

$\displaystyle \textbf{Question 18: }\text{Sketch the graph }y=\lvert x+3\rvert.\ \text{Evaluate }\int_{-6}^{0}\lvert x+3\rvert\,dx.\ \\ \text{What does this integral represent on the graph?}$
$\displaystyle \text{Answer:}$

$\displaystyle y=|x+3|\text{ intersects }x=0\text{ and }x=-6\text{ at }(0,3)\text{ and }(-6,3)\text{ respectively}$
$\displaystyle \text{Now}$
$\displaystyle y=|x+3|$
$\displaystyle =\begin{cases}x+3,&x>-3\\-(x+3),&x<-3\end{cases}$
$\displaystyle \text{Integral represents the area enclosed between }x=-6\text{ and }x=0$
$\displaystyle A=\int_{-6}^{0}|y|\,dx$
$\displaystyle =\int_{-6}^{-3}|y|\,dx+\int_{-3}^{0}|y|\,dx$
$\displaystyle =\int_{-6}^{-3}-(x+3)\,dx+\int_{-3}^{0}(x+3)\,dx$
$\displaystyle =-\left[\frac{x^2}{2}+3x\right]_{-6}^{-3}+\left[\frac{x^2}{2}+3x\right]_{-3}^{0}$
$\displaystyle =-\left[\left(\frac{9}{2}-9\right)-\left(18-18\right)\right]+\left[\left(0+0\right)-\left(\frac{9}{2}-9\right)\right]$
$\displaystyle =-\left(-\frac{9}{2}\right)+\frac{9}{2}$
$\displaystyle =9$
$\displaystyle \therefore \text{Area enclosed }=9\text{ sq. units}$

$\displaystyle \textbf{Question 19: }\text{Sketch the graph }y=\lvert x+1\rvert.\ \text{Evaluate }\int_{-4}^{2}\lvert x+1\rvert\,dx.\ \\ \text{What does the value of this integral represent on the graph?}$
$\displaystyle \text{Answer:}$

$\displaystyle y=|x+1|\text{ intersects }x=-4\text{ and }x=2\text{ at }(-4,3)\text{ and }(2,3)\text{ respectively}$
$\displaystyle \text{Now}$
$\displaystyle y=|x+1|$
$\displaystyle =\begin{cases}x+1,&x>-1\\-(x+1),&x<-1\end{cases}$
$\displaystyle \text{Integral represents the area enclosed between }x=-4\text{ and }x=2$
$\displaystyle A=\int_{-4}^{2}|y|\,dx$
$\displaystyle =\int_{-4}^{-1}|y|\,dx+\int_{-1}^{2}|y|\,dx$
$\displaystyle =\int_{-4}^{-1}-(x+1)\,dx+\int_{-1}^{2}(x+1)\,dx$
$\displaystyle =-\left[\frac{x^{2}}{2}+x\right]_{-4}^{-1}+\left[\frac{x^{2}}{2}+x\right]_{-1}^{2}$
$\displaystyle =-\left[\left(\frac{1}{2}-1\right)-\left(8-4\right)\right]+\left[\left(2+2\right)-\left(\frac{1}{2}-1\right)\right]$
$\displaystyle =\left(3-\frac{15}{2}\right)+\left(5-\frac{1}{2}\right)$
$\displaystyle =-3+\frac{15}{2}+5-\frac{1}{2}$
$\displaystyle =9$
$\displaystyle \therefore \text{Area enclosed }=9\text{ sq. units}$

$\displaystyle \textbf{Question 20: }\text{Find the area of the region bounded by the curve } \\ xy-3x-2y-10=0,\ x\text{-axis and the lines }x=3,\ x=4.$
$\displaystyle \text{Answer:}$

$\displaystyle \text{We have}$
$\displaystyle xy-3x-2y-10=0$
$\displaystyle \Rightarrow xy-2y=3x+10$
$\displaystyle \Rightarrow y(x-2)=3x+10$
$\displaystyle \Rightarrow y=\frac{3x+10}{x-2}$
$\displaystyle \text{Let }A\text{ represent the required area}$
$\displaystyle A=\int_3^4 |y|\,dx$
$\displaystyle =\int_3^4 \frac{3x+10}{x-2}\,dx$
$\displaystyle =\int_3^4 \frac{3x-6+16}{x-2}\,dx$
$\displaystyle =\int_3^4 \left(3+\frac{16}{x-2}\right)\,dx$
$\displaystyle =\left[3x+16\log|x-2|\right]_3^4$
$\displaystyle =\left[12+16\log 2\right]-\left[9+16\log 1\right]$
$\displaystyle =3+16\log 2$
$\displaystyle \therefore \text{Required area }=3+16\log 2\text{ sq. units}$

$\displaystyle \textbf{Question 21: }\text{Draw a rough sketch of the curve }y=\frac{\pi}{2}+2\sin^{2}x \text{ and find } \\ \text{the area between }x\text{-axis, the curve and the ordinates }x=0,\ x=\pi.$
$\displaystyle \text{Answer:}$

$\displaystyle \begin{array}{|c|c|c|c|c|c|} \hline x & 0 & \frac{\pi}{6} & \frac{\pi}{2} & \frac{5\pi}{6} & \pi \\ \hline \sin x & 0 & \frac{1}{2} & 1 & \frac{1}{2} & 0 \\ \hline y=\frac{\pi}{2}+2\sin^2 x & 1.57 & 2.07 & 3.57 & 2.07 & 1.57 \\ \hline \end{array}$
$\displaystyle \text{Given }y=\frac{\pi}{2}+2\sin^2 x$
$\displaystyle \text{The curve cuts the }y\text{-axis at }(0,\tfrac{\pi}{2})\text{ and at }x=\pi\text{ at }(\pi,\tfrac{\pi}{2})$
$\displaystyle x=\pi\text{ is a line parallel to the }y\text{-axis}$
$\displaystyle \text{Consider a vertical strip of length }|y|\text{ and width }dx\text{ in the first quadrant}$
$\displaystyle \text{Area of the approximating rectangle }=|y|\,dx$
$\displaystyle \text{The approximating rectangle moves from }x=0\text{ to }x=\pi$
$\displaystyle \therefore \text{Area of the shaded region }=\int_0^{\pi}|y|\,dx$
$\displaystyle =\int_0^{\pi} y\,dx \text{ since }y>0$
$\displaystyle =\int_0^{\pi}\left(\frac{\pi}{2}+2\sin^2 x\right)dx$
$\displaystyle =\int_0^{\pi}\left(\frac{\pi}{2}+2\cdot\frac{1-\cos 2x}{2}\right)dx$
$\displaystyle =\int_0^{\pi}\left(\frac{\pi}{2}+1-\cos 2x\right)dx$
$\displaystyle =\frac{\pi}{2}\int_0^{\pi}dx+\int_0^{\pi}(1-\cos 2x)\,dx$
$\displaystyle =\frac{\pi}{2}[x]_0^{\pi}+\left[x-\frac{\sin 2x}{2}\right]_0^{\pi}$
$\displaystyle =\frac{\pi}{2}(\pi)+\left(\pi-0\right)$
$\displaystyle =\frac{\pi^2}{2}+\pi$
$\displaystyle =\frac{\pi}{2}(\pi+2)$
$\displaystyle \therefore \text{Area of the curve bounded by }x=0\text{ and }x=\pi\text{ is }\frac{\pi}{2}(\pi+2)\text{ sq. units}$

$\displaystyle \textbf{Question 22: }\text{Draw a rough sketch of the curve }y=\frac{x}{\pi}+2\sin^{2}x \text{ and find } \\ \text{the area between }x\text{-axis, the curve and the ordinates }x=0\text{ and }x=\pi.$
$\displaystyle \text{Answer:}$

$\displaystyle \text{The table for different values of }x\text{ and }y\text{ is}$
$\displaystyle \begin{array}{|c|c|c|c|c|c|} \hline x & 0 & \frac{\pi}{6} & \frac{\pi}{2} & \frac{5\pi}{6} & \pi \\ \hline \sin x & 0 & \frac{1}{2} & 1 & \frac{1}{2} & 0 \\ \hline y=\frac{x}{\pi}+2\sin^2 x & 0 & \frac{2}{3} & \frac{5}{2} & \frac{4}{3} & 1 \\ \hline \end{array}$
$\displaystyle y=\frac{x}{\pi}+2\sin^2 x \text{ is an arc cutting the }y\text{-axis at }O(0,0)\text{ and cutting }x=\pi\text{ at }(\pi,1)$
$\displaystyle \text{Consider a vertical strip of length }|y|\text{ and width }dx\text{ in the first quadrant}$
$\displaystyle \text{Area of the approximating rectangle }=|y|\,dx$
$\displaystyle \text{The approximating rectangle moves from }x=0\text{ to }x=\pi$
$\displaystyle \therefore \text{Area of the shaded region }=\int_0^{\pi}|y|\,dx$
$\displaystyle =\int_0^{\pi} y\,dx \text{ since } y>0 \Rightarrow |y|=y$
$\displaystyle =\int_0^{\pi}\left(\frac{x}{\pi}+2\sin^2 x\right)dx$
$\displaystyle =\frac{1}{\pi}\int_0^{\pi}x\,dx+2\int_0^{\pi}\sin^2 x\,dx$
$\displaystyle =\frac{1}{\pi}\left[\frac{x^2}{2}\right]_0^{\pi}+2\left[\frac{x}{2}-\frac{\sin x\cos x}{2}\right]_0^{\pi}$
$\displaystyle =\frac{\pi^2}{2\pi}+2\left[\frac{\pi}{2}-0\right]$
$\displaystyle =\frac{\pi}{2}+\pi$
$\displaystyle =\frac{3\pi}{2}$
$\displaystyle \therefore \text{Area of the curve enclosed between }x=0\text{ and }x=\pi\text{ is }\frac{3\pi}{2}\text{ sq. units}$

$\displaystyle \textbf{Question 23: }\text{Find the area bounded by the curve }y=\cos x,\ x\text{-axis and the } \\ \text{ordinates } x=0\text{ and }x=2\pi.$
$\displaystyle \text{Answer:}$

$\displaystyle \text{The shaded region is the required area bounded by the curve }y=\cos x,\ x\text{-axis and }x=0,\ x=2\pi$
$\displaystyle \text{Consider a vertical strip of length }|y|\text{ and width }dx$
$\displaystyle \text{Area of the approximating rectangle }=|y|\,dx$
$\displaystyle \text{The approximating rectangle moves from }x=0\text{ to }x=2\pi$
$\displaystyle \text{Now, }0\le x\le\frac{\pi}{2}\text{ and }\frac{3\pi}{2}\le x\le2\pi,\ y>0\Rightarrow |y|=y$
$\displaystyle \frac{\pi}{2}\le x\le\frac{3\pi}{2},\ y<0\Rightarrow |y|=-y$
$\displaystyle \therefore \text{Area of the shaded region }=\int_0^{2\pi}|y|\,dx$
$\displaystyle =\int_0^{\pi/2}|y|\,dx+\int_{\pi/2}^{3\pi/2}|y|\,dx+\int_{3\pi/2}^{2\pi}|y|\,dx$
$\displaystyle =\int_0^{\pi/2}y\,dx+\int_{\pi/2}^{3\pi/2}(-y)\,dx+\int_{3\pi/2}^{2\pi}y\,dx$
$\displaystyle =\int_0^{\pi/2}\cos x\,dx-\int_{\pi/2}^{3\pi/2}\cos x\,dx+\int_{3\pi/2}^{2\pi}\cos x\,dx$
$\displaystyle =[\sin x]_0^{\pi/2}-[\sin x]_{\pi/2}^{3\pi/2}+[\sin x]_{3\pi/2}^{2\pi}$
$\displaystyle =1-( -1-1)+(0-(-1))$
$\displaystyle =4$
$\displaystyle \therefore \text{Area bounded by the curve }y=\cos x,\ x\text{-axis and }x=0,\ x=2\pi\text{ is }4\text{ sq. units}$

$\displaystyle \textbf{Question 24: }\text{Show that the areas under the curves }y=\sin x\text{ and }y=\sin 2x \\ \text{ between } x=0\text{ and }x=\frac{\pi}{3}\text{ are in the ratio }2:3.$
$\displaystyle \text{Answer:}$

$\displaystyle \text{The shaded area }A_1\text{ is the required area bounded by the curve }y=\sin x\text{ and }x=0,\ x=\frac{\pi}{3}$
$\displaystyle \text{The shaded area }A_2\text{ is the required area bounded by the curve }y=\sin 2x\text{ and }x=0,\ x=\frac{\pi}{3}$
$\displaystyle A_1=\int_0^{\pi/3}|y|\,dx$
$\displaystyle =\int_0^{\pi/3}y\,dx \text{ since }y>0\Rightarrow |y|=y$
$\displaystyle =\int_0^{\pi/3}\sin x\,dx$
$\displaystyle =[-\cos x]_0^{\pi/3}$
$\displaystyle =-\cos\frac{\pi}{3}+\cos 0$
$\displaystyle =-\frac{1}{2}+1=\frac{1}{2}\quad\text{(1)}$
$\displaystyle A_2=\int_0^{\pi/3}|y|\,dx$
$\displaystyle =\int_0^{\pi/3}y\,dx \text{ since }y>0\Rightarrow |y|=y$
$\displaystyle =\int_0^{\pi/3}\sin 2x\,dx$
$\displaystyle =\int_0^{\pi/3}2\sin x\cos x\,dx$
$\displaystyle =2\int_0^{\pi/3}\sin x\cos x\,dx$
$\displaystyle =2\left[-\frac{1}{4}\cos 2x\right]_0^{\pi/3}$
$\displaystyle =\frac{1}{2}\left[-\cos\frac{2\pi}{3}+\cos 0\right]$
$\displaystyle =\frac{1}{2}\left[\frac{1}{2}+1\right]=\frac{3}{4}\quad\text{(2)}$
$\displaystyle \text{From (1) and (2)}$
$\displaystyle \frac{A_1}{A_2}=\frac{\tfrac{1}{2}}{\tfrac{3}{4}}=\frac{2}{3}$
$\displaystyle \therefore \text{The areas bounded by }y=\sin x\text{ and }y=\sin 2x\text{ for }x=0\text{ to }x=\frac{\pi}{3}\text{ are in the ratio }2:3$

$\displaystyle \textbf{Question 25: }\text{Compare the areas under the curves }y=\cos^{2}x\text{ and } y=\sin^{2}x \\ \text{ between }x=0\text{ and }x=\pi.$
$\displaystyle \text{Answer:}$

$\displaystyle \text{The table for different values of }x\text{ is}$
$\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|} \hline x & 0 & \frac{\pi}{4} & \frac{\pi}{3} & \frac{\pi}{2} & \frac{2\pi}{3} & \frac{5\pi}{6} & \pi \\ \hline y=\cos^2 x & 1 & \frac{1}{2} & \frac{1}{4} & 0 & \frac{1}{4} & \frac{3}{4} & 1 \\ \hline y=\sin^2 x & 0 & \frac{1}{2} & \frac{3}{4} & 1 & \frac{3}{4} & \frac{1}{4} & 0 \\ \hline \end{array}$
$\displaystyle \text{Let }A_1\text{ be the area of the curve }y=\cos^2 x\text{ between }x=0\text{ and }x=\pi$
$\displaystyle \text{Let }A_2\text{ be the area of the curve }y=\sin^2 x\text{ between }x=0\text{ and }x=\pi$
$\displaystyle \text{Consider a vertical strip of length }|y|\text{ and width }dx$
$\displaystyle \text{Area of the approximating rectangle }=|y|\,dx$
$\displaystyle \text{The approximating rectangle moves from }x=0\text{ to }x=\pi$
$\displaystyle A_1=\int_0^{\pi}|y|\,dx$
$\displaystyle =\int_0^{\pi}y\,dx \text{ since }0\le x\le\pi,\ y\ge0\Rightarrow |y|=y$
$\displaystyle =\int_0^{\pi}\cos^2 x\,dx$
$\displaystyle =\int_0^{\pi}\frac{1+\cos 2x}{2}\,dx$
$\displaystyle =\frac{1}{2}\left[x+\frac{\sin 2x}{2}\right]_0^{\pi}$
$\displaystyle =\frac{1}{2}\left[\pi+0\right]$
$\displaystyle =\frac{\pi}{2}\text{ sq. units}$
$\displaystyle A_2=\int_0^{\pi}|y|\,dx$
$\displaystyle =\int_0^{\pi}y\,dx \text{ since }0\le x\le\pi,\ y\ge0\Rightarrow |y|=y$
$\displaystyle =\int_0^{\pi}\sin^2 x\,dx$
$\displaystyle =\int_0^{\pi}\frac{1-\cos 2x}{2}\,dx$
$\displaystyle =\frac{1}{2}\left[x-\frac{\sin 2x}{2}\right]_0^{\pi}$
$\displaystyle =\frac{1}{2}\left[\pi-0\right]$
$\displaystyle =\frac{\pi}{2}\text{ sq. units}$
$\displaystyle \therefore \text{Areas of the curves }y=\cos^2 x\text{ and }y=\sin^2 x\text{ between }x=0\text{ and }x=\pi\text{ are equal and each is }\frac{\pi}{2}\text{ sq. units}$

$\displaystyle \textbf{Question 26: }\text{Find the area bounded by the ellipse }\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \\ \text{ and the ordinates }x=ae\text{ and }x=0,\ \text{where }b^{2}=a^{2}(1-e^{2})\text{ and }e<1.$
$\displaystyle \text{Answer:}$

$\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \text{ represents an ellipse, symmetrical about both the axes.}$
$\displaystyle \text{It cuts the } x\text{-axis at } A(a,0) \text{ and } A'(-a,0).$
$\displaystyle \text{It cuts the } y\text{-axis at } B(0,b) \text{ and } B'(0,-b).$
$\displaystyle x=ae \text{ is a line parallel to the } y\text{-axis.}$
$\displaystyle \text{Consider a vertical strip of length } |y| \text{ and width } dx \text{ in the first quadrant.}$
$\displaystyle \text{Area of approximating rectangle in first quadrant } = |y|\,dx.$
$\displaystyle \text{Approximating rectangle moves from } x=0 \text{ to } x=ae.$
$\displaystyle \text{Area of the shaded region } = 2 \times \text{area in the first quadrant.}$
$\displaystyle \Rightarrow A = 2\int_0^{ae} |y|\,dx.$
$\displaystyle \Rightarrow A = 2\int_0^{ae} y\,dx \quad (\text{as } y>0,\ |y|=y).$
$\displaystyle \Rightarrow A = 2\int_0^{ae} \frac{b}{a}\sqrt{a^2-x^2}\,dx \quad \left(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \Rightarrow y=\frac{b}{a}\sqrt{a^2-x^2}\right).$
$\displaystyle \Rightarrow A = \frac{2b}{a}\int_0^{ae}\sqrt{a^2-x^2}\,dx.$
$\displaystyle \Rightarrow A = \frac{2b}{a}\left[\frac{1}{2}x\sqrt{a^2-x^2}+\frac{1}{2}a^2\sin^{-1}\!\left(\frac{x}{a}\right)\right]_0^{ae}.$
$\displaystyle \Rightarrow A = \frac{2b}{a}\left[\frac{1}{2}ae\sqrt{a^2-a^2e^2}+\frac{1}{2}a^2\sin^{-1}(e)\right].$
$\displaystyle \Rightarrow A = \frac{2b}{a}\left[\frac{1}{2}a^2 e\sqrt{1-e^2}+\frac{1}{2}a^2\sin^{-1}(e)\right].$
$\displaystyle \Rightarrow A = ab\left[e\sqrt{1-e^2}+\sin^{-1}(e)\right].$
$\displaystyle \therefore \text{Required area of the ellipse bounded by } x=0 \text{ and } x=ae \text{ is } ab\left[e\sqrt{1-e^2}+\sin^{-1}(e)\right] \text{ sq. units.}$

$\displaystyle \textbf{Question 27: }\text{Find the area of the minor segment of the circle }x^{2}+y^{2}=a^{2} \\ \text{ cut off by the line }x=\frac{a}{2}.$
$\displaystyle \text{Answer:}$

$\displaystyle \text{The equation of the circle is } x^2+y^2=a^2.$
$\displaystyle \text{Centre of the circle is } (0,0) \text{ and radius is } a.$
$\displaystyle \text{The line } x=\frac{a}{2} \text{ is parallel to the } y\text{-axis and intersects the } x\text{-axis at } \left(\frac{a}{2},0\right).$
$\displaystyle \text{Required area } = \text{Area of the shaded region.}$
$\displaystyle = 2 \times \text{Area of the region } ABDA.$
$\displaystyle = 2\int_{a/2}^{a} y\,dx.$
$\displaystyle = 2\int_{a/2}^{a} \sqrt{a^2-x^2}\,dx.$
$\displaystyle = 2\left[\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\!\left(\frac{x}{a}\right)\right]_{a/2}^{a}.$
$\displaystyle = 2\left[\left(0+\frac{a^2}{2}\sin^{-1}(1)\right)-\left(\frac{a}{4}\sqrt{a^2-\frac{a^2}{4}}+\frac{a^2}{2}\sin^{-1}\!\left(\frac{1}{2}\right)\right)\right].$
$\displaystyle = 2\left[\frac{a^2}{2}\cdot\frac{\pi}{2}-\frac{a}{4}\cdot\frac{\sqrt{3}a}{2}-\frac{a^2}{2}\cdot\frac{\pi}{6}\right].$
$\displaystyle = \frac{a^2\pi}{2}-\frac{\sqrt{3}a^2}{4}-\frac{a^2\pi}{6}.$
$\displaystyle = \frac{6a^2\pi-3\sqrt{3}a^2-2a^2\pi}{12}.$
$\displaystyle = \frac{a^2}{12}\left(4\pi-3\sqrt{3}\right) \text{ square units.}$

$\displaystyle \textbf{Question 28: }\text{Find the area of the region bounded by the curve } \\ x=at^{2},\ y=2at \text{ between the ordinates corresponding }t=1\text{ and }t=2.$
$\displaystyle \text{Answer:}$

$\displaystyle \text{The curve } x=at^2,\ y=2at \text{ represents the parametric equation of a parabola.}$
$\displaystyle \text{Eliminating the parameter } t,\ \text{we get } y^2=4ax.$
$\displaystyle \text{This represents the Cartesian equation of a parabola opening towards the positive } x\text{-axis with focus at } (a,0).$
$\displaystyle \text{When } t=1,\ x=a.$
$\displaystyle \text{When } t=2,\ x=4a.$
$\displaystyle \therefore \text{Required area } = \text{Area of the shaded region.}$
$\displaystyle = 2 \times \text{Area of the region } ABCFA.$
$\displaystyle = 2\int_a^{4a} y\,dx.$
$\displaystyle = 2\int_a^{4a} \sqrt{4ax}\,dx.$
$\displaystyle = 2\int_a^{4a} 2\sqrt{ax}\,dx.$
$\displaystyle = 4\sqrt{a}\int_a^{4a} x^{1/2}\,dx.$
$\displaystyle = 4\sqrt{a}\left[\frac{2}{3}x^{3/2}\right]_a^{4a}.$
$\displaystyle = \frac{8\sqrt{a}}{3}\left[(4a)^{3/2}-a^{3/2}\right].$
$\displaystyle = \frac{8\sqrt{a}}{3}\left[8a\sqrt{a}-a\sqrt{a}\right].$
$\displaystyle = \frac{8\sqrt{a}}{3}\cdot 7a\sqrt{a}.$
$\displaystyle = \frac{56}{3}a^2 \text{ square units.}$

$\displaystyle \textbf{Question 29: }\text{Find the area enclosed by the curve }x=3\cos t,\ y=2\sin t.$
$\displaystyle \text{Answer:}$

$\displaystyle \text{The given curve } x=3\cos t,\ y=2\sin t \text{ represents the parametric equation of an ellipse.}$
$\displaystyle \text{Eliminating the parameter } t,\ \text{we get}$
$\displaystyle \frac{x^2}{9}+\frac{y^2}{4}=\cos^2 t+\sin^2 t=1.$
$\displaystyle \text{This represents the Cartesian equation of the ellipse with centre } (0,0).$
$\displaystyle \text{The coordinates of the vertices are } (\pm 3,0) \text{ and } (0,\pm 2).$
$\displaystyle \therefore \text{Required area } = \text{Area of the shaded region.}$
$\displaystyle = 4 \times \text{Area of the region } OABO.$
$\displaystyle = 4\int_0^{3} y\,dx.$
$\displaystyle = 4\int_0^{3} 2\sqrt{1-\frac{x^2}{9}}\,dx.$
$\displaystyle = \frac{8}{3}\int_0^{3} \sqrt{9-x^2}\,dx.$
$\displaystyle = \frac{8}{3}\left[\frac{x}{2}\sqrt{9-x^2}+\frac{9}{2}\sin^{-1}\!\left(\frac{x}{3}\right)\right]_0^{3}.$
$\displaystyle = \frac{8}{3}\left[\left(0+\frac{9}{2}\sin^{-1}(1)\right)-(0+0)\right].$
$\displaystyle = \frac{8}{3}\cdot\frac{9}{2}\cdot\frac{\pi}{2}.$
$\displaystyle = 6\pi \text{ square units.}$